Geometry and Kinematic Singularities of

closed-loop manipulators

Jean-Pierre MERLET ∗

Received: November 27, 2013

∗INRIA, Centre de Sophia Antipolis, 2004 Route des Lucioles,06565 Valbonne, France, E-mail:

merlet@cygnusx1.inria.fr

1

Abstract

The determination of the singularities of closed-loop manipulators is in general a difficultproblem. We consider in this paper specific classes of closed-loop manipulators and we showhow we can define a singular configuration. The behaviour of the robot in the vicinity ofthese configurations is examined. We describe then a geometrical approach which enablesto find the loci of the singular configurations together with a geometrical description of therobot in these configurations. This approach is illustrated by the complete analysis of agiven manipulator.

1 Parallel manipulator

1.1 Introduction

Parallel manipulators are closed-loop mechanism in which all the links areconnected both at the base and at the gripper of the robot. Manipulatorsof this type have been designed or studied for a long time. The first one,to the author’s knowledge, was designed for testing tyres (see Mc Gough inStewart paper [20]). But this mechanical architecture is mainly used for theflight simulator (see for example Stewart [20], Baret [2]). The first design asa manipulator system has been done by Mac Callion in 1979 for an assemblyworkstation [13]. Some other researchers have also addressed this problem:Arai [1], Fichter [6], Gosselin [7], Herve [8],Inoue [12], Reboulet [19], Yang[23], Zamanov [24].

To illustrate our approach we will consider a specific mechanical architecturecalled a SSM described in Figure 1. Basically it consists in two plates connectedby 6 articulated links. In the following sections the smaller plate will be calledthe mobile and the larger ( which is in general fixed) will be called the base. Ineach articulated link there is one linear actuator and by changing the lengthsof the links we are able to control the position 1 of the gripper. The SSM canbe simplified if the hexagonal mobile plate is changed to a triangular plate (wewill call this kind of manipulator a TSSM). A further simplification is obtainedwhen both plates are triangles and the resulting manipulator is called a MSSM(Figure 8).

1In this paper position means position and orientation

2

SSMC

y

x

A1

A2

A3

A4

A5

A6

B1

B2

B3

B4

B5

B6

A2

A1

A4

A6

A3

A5

mobile

base

1

2

3

4

5

6

O

C

Figure 1: A parallel manipulator: the SSM. Links i is articulated at point Ai, Bi. (perspec-tive and top view)

1.2 Notation

We introduce an absolute frame R with origin O and a relative frame Rb fixedto the mobile with origin C (see Figure 1). The rotation matrix relating avector in Rb to the same vector in R will be denoted by M .

The center of the articulations on the base for link i will be denoted Ai

and that on the mobile Bi. The length of link i will be noted ρi, and the unitvector of this link ni. The coordinates of Ai in frame R are (xai, yai, zai),the coordinates of Bi in frame Rb are (xi, yi, zi) and the coordinates of C,the origin of the relative frame, Xc = (xc, yc, zc). We use the Euler’s anglesΩc = (ψ, θ, φ) to represent the orientation of the mobile.

For the sake of simplicity the subscript i is omitted whenever it is possi-ble and vectors will be noted in bold character. A vector with coordinatesexpressed in the relative frame will be denoted by the subscript r.

We will consider the case where each set of articulation points of both thebase and the mobile lie in a plane. In this case, without loss of generality, wewill define R such that zai = 0 and Rb such that zi = 0 .

1.3 Inverse and Direct kinematics

Let us calculate the fundamental relations between the links lengths and theposition of the mobile. For a given link we have :

AB = ρn AB = AO + OC + CB CB = MCBr (1)

where CBr means the coordinates of the articulation points with respect tothe frame Rb. n being a unit vector we have :

ρ = ||AO + OC+MCBr|| = ||U|| (2)

3

If the position of the mobile is given we are able to calculate the componentsof U and thus the length of the segment. Therefore the inverse kinematics isstraightforward (this is in fact a general feature of parallel manipulators) andis defined by the above 6 equations which constitute a system of non-linearequations denoted by S.

At the opposite the direct kinematics is much more complicated. Indeed tofind the position of the mobile for a given set of links lengths we have to solvethe system S. It has been shown in [17] that in general the solution is notunique : if the mobile plate is a triangle up to 16 solutions can exist and inthe case of the SSM it has been shown that an upperbound of the number ofsolution is 352 although a numerical study has yield to at most 12 solutions.

2 Singularities

2.1 An analytical approach

Let us assume that for a given set of links lengths ρ we know a solution X0 ofthe system S. From the rank theorem we know that in a neighborhood of X0

the solution of S is unique if the rank of the jacobian matrix J of this systemis equal to 6 with:

J = ((∂ρ

∂X)) (3)

where X is the position parameters vector. Note that this matrix is in fact theinverse jacobian (in a robotics sense) of the manipulator. Now let us assumethat J is singular : this means that the mobile plate may have an infinitesimal

motion around X0 without any change in the links lengths. In that case we willsay that X0 is a singular configuration of the manipulator. In other words thevelocity of the mobile plate may be different from zero although the actuator’svelocities are all equal to zero. This means also that in these configurationsthe manipulator gains some degrees of freedom (at the opposite of the singularconfigurations of serial manipulator where it loses degrees of freedom).

2.2 A mechanical approach

The previous approach indicates that in a singular configuration the manip-ulator is no more controllable. But a mechanical approach will give anotherinsight of these configurations. Let τ denotes the articular forces vector (i.e.the traction-compression stress in the links) and F an external wrench appliedon the mobile plate. It is well known that we have:

F = JT τ (4)

4

12

36

4,5

3,61,2

12

B5B3B3, B5

B1

B1

Figure 2: Hunt’s singular configuration for the TSSM.

If J is singular then no τ can be found to equilibrate a set of wrench. Further-more in the vicinity of a singular configuration the articular forces will tendto infinity. Practically this imply that if the mobile plate is ”close” from asingular configuration the robot will suffer mechanical damages and this ex-plain why the determination of the loci of these configurations is an importantproblem.

2.3 The determinant of J

From this point the solution to this problem seems obvious : as the matrix Jis completely determined we calculate its determinant and find its roots in X.In fact the symbolic computation of this determinant is rather tedious ( see[14] for the formulation of this determinant). To get an idea of its complexitythe computation of the determinant of a SSM involves 29 powers , 21530multiplications, 915 additions and 907 subtractions.....

2.4 Previous works

Few researchers have addressed the problem of determining the loci of thesingular configurations.

Using a mechanical analysis Hunt [10] has determined a singular configura-tion for a TSSM (Figure 2). In this configuration all the segments intersect oneline ( line B3B5) and an external torque around this line cannot be equilibratedby the actuator forces.

Fichter [6] describes another singular configuration which is obtained whenthe mobile plate is rotated around the z axis with an angle of ±π

2. This config-

uration was obtained by noticing that in this case two lines of the determinant

5

X

Y

Z

M

M1

M2

S

O

Figure 3: Plucker coordinates.

were constant. But outside these two particular configurations no system-atic method was proposed to find all the singular configurations of a parallelmanipulator.

3 A geometrical approach for finding the singularities

3.1 Plucker coordinates of lines

It is well known that a line can be described by its Plucker coordinates. Letus introduce briefly these coordinates. We consider two points on a line, sayM1 and M2, and a reference frame R0 whose origin is O (see Figure 3).

Let us consider now the two three dimensional vectors S and M defined by:

S = M1M2 M = OM1 ∧ OM2 = OM2 ∧ S = OM1 ∧ S

If we assemble these vectors to form a six-dimensional vector we get the Pluckervector Lp of this line.

Lp = [Sx, Sy, Sz,Mx,My,Mz]

Let us assume now that the Plucker vectors belong to a vector space V6 andwe consider the one-dimensional subspaces of V6 as points of a projective P5.Then every line g in P3 corresponds to exactly one point G in P5.

It is well known that point G belongs to a quadric Qp (see [4], [22], [3]).Indeed we have for every line of P3 :

SxMx + SyMy + SzMz = 0

This equation defines the quadric Qp which is called the Grassmannian or thePlucker quadric. At this point we have defined a one-to-one relation between

6

the set of lines in the real P3 and the quadric Qp in P5. The rank of thismapping is 6 (there is at most 6 independent Plucker vectors).

Let us consider now the various sub-spaces of P5 (or more precisely theirintersection with Qp). We get various varieties whose rank ranges from 0 to6. As a matter of example a point in P5 ( rank=1) corresponds to a line inP3. As for Qp (which represents the set of line of P3) it is defined through 6linearly independent Plucker vectors and is therefore of rank 6.

3.2 Plucker vectors of the links and matrix J

Consider the equilibrium of the mobile plate under the effect of an externalwrench T = (F,M) and the actuators force vector τ . The equilibrium condi-tions are:

F =i=6∑

i=1

τini M =i=6∑

i=1

CBi ∧ τini (5)

which can be written as :

T = ((ni CBi ∧ ni))T τ (6)

Using equation (4) we get :

J = ((ni CBi ∧ ni)) (7)

and therefore row i of matrix J is equal to the Plucker vector of the lineassociated to link i. Although we have obtained this result for a SSM it canbe extended to various kind of parallel manipulator [18] (even for manipulatorwith less than 6 degrees of freedom).

From this results we deduce that a degeneracy of matrix J imply a lineardependence between the six 6-dimensional Plucker vectors of the line associatedto the links or in other words that the variety spanned by these lines has arank less than 6.

4 Grassmann Geometry

The varieties spanned by a set of lines has been studied by H. Grassmann(1809-1877). The purpose of his study was to find geometric characterizationsof each varieties i.e. find all the geometric conditions on a set of m lines suchthat these lines spanned a variety of rank n with n < m ≤ 6. We will introducenow the various results which can be found in [5] or, with more mathematicaljustifications, in [22].

Let us begin with the linear varieties of rank 0 through 3 (Figure 4 ). Wehave first the empty set of rank 0. Then the point (rank=1), which is a line

7

3

2

1

rang

a) b)

c)d)

Figure 4: Grassmann varieties of rank 1,2,3.

in the 3D space. The lines (rank=2) are either a pair of skew lines in R3 or aflat pencil of lines: those lying in a plane and passing through some point onthat plane.

The planes (rank=3) are of four types:

• all lines in a plane (3d)

• all lines through a point (3c)

• the union of two flat pencils having a line in common but lying in distinctsplanes and with distinct centers (3b)

• a regulus (3a)

Let us define the regulus. Take three skew lines in space and consider the setof lines which intersect these three lines : this set of lines build a surface whichis an hyperboloid of one sheet (a quadric surface) and is called a regulus.Each line belonging to the regulus is called a generator of the regulus. It isshown in [9],[22] that this surface is doubly ruled. This means that there existtwo reguli (a regulus and its ”complementary” regulus) which generate thesame surface or that each point on the surface is on more than one line.

Therefore there are two families of straight lines on the hyperboloid andeach family covers the surface completely. A line on this surface is dependenton the lines of either the regulus or the complementary regulus. An interesting

8

4a

4b

4c

4d 5a

5b

Figure 5: Grassmann varieties of rank 4,5.

property is that a line of one family intersects all the lines of the other familyand that any two lines of the same family are mutually skew (see [21] for thehairy details).

Let us describe now the linear varieties of higher rank of the Grassmanngeometry (Figure 5 ). Linear varieties of dimension 4 are called linear congru-

ences and are of four types:

• a linear spread generated by four skew lines i.e. no line meet the regulusgenerated by the three others lines in a proper point (elliptic congruence,4a)

• all the lines concurrent with two skew lines (hyperbolic congruence, 4b)

• a one-parameter family of flat pencil, having one line in common andforming a variety (parabolic congruence, 4c)

• all the lines in a plane or passing through one point in that plane (degen-

erate congruence, 4d)

Linear varieties of dimension 5 are called linear complexes and are of twotypes:

• non singular (or general): generated by five independent skew lines (5a)

• singular (or special): all the lines meeting one given line (5b)

The geometric characterization of a general linear complex is that throughany point of the space there is one and only one flat pencil of line such thatall the lines which belong to the pencil belong also to the complex. In otherwords all the lines of a linear complex which are coplanar intersect one point.

9

mobile

articulation

link

base

y

x

1

2

3

B1(xb1, yb1, 0)

B2(xb2, yb2, 0)

B3(xb3, yb3, 0)

Figure 6: A 2D parallel manipulator. By changing the lengths of links (1,2,3) the positionand orientation of the mobile can be controlled.

5 Application of Grassmann geometry to the determi-

nation of the singularities

If we consider the example of a SSM in a singular configuration the rank ofthe variety spanned by the lines associated to the links will be less than 6.This imply that it exists at least a set of m(m ≤ 6) lines which spanned avariety of rank m− 1. Therefore to find the position of the mobile plate suchthat the SSM is in a singular configuration we will examine each set of mlines (m ∈ [3, 6]) and determine the position of the mobile plate such that thegeometric condition given by Grassmann geometry for a variety of rank m− 1is fulfilled. For example we will consider all the set of 4 lines and determine theposition of the mobile plate for which these 4 lines have a common point: inthis case the rank of the variety spanned by the four lines is 3 and the matrixJ is singular.

5.1 A basic example: the 2D parallel manipulator

Let us consider a basic example: a 2D parallel manipulator (Figure 6 ). Theequilibrium condition can be written in matrix form as:

Fx

Fy

Fz

=

n1xn2x

n3x

n1yn2y

n3y

n1yxb1 − yb1n1x

n2yxb2 − yb2n2x

n3yxb3 − yb3n3x

τ1τ2τ3

(8)Let us consider the three column vectors Ti of the above 3x3 matrix. If one ofthem is linearly dependent from the two others then the manipulator is in a

10

singular configuration. For each of these vectors we may build an augmented6 dimensional vector Si by adding three 0 :

Si = [nix , niy , 0, 0, 0, niyxbi − ybinix ]

Clearly if one the Ti is linearly dependent from the two others then the cor-responding Si will also be linearly dependent from the two others and theopposite is also true. It must then be noticed that the Si is the Plucker vectorof the line associated to link i. Therefore in this case we have three Pluckervectors and we are looking for a configuration of the manipulator such thatthe rank of the variety spanned by these 3 lines is 2.

By reference to Figure 4 we can see that the only possibility for a systemof three coplanar bars to be a 2-rank Grassmann variety is obtained when thethree lines cross the same point (Figure 7). Therefore the loci of the singular

Figure 7: Singular configuration for the 2D parallel manipulator : the three line associatedto the links have a common intersection point.

configurations expressed as position of the mobile plate can be easily describedfrom a geometrical view point.

6 Study of the MSSM

We will deal now with a more complete example of a 6 d.o.f. manipulatorcalled the MSSM (Figure 8). A previous analysis [15] has shown that amongthe Grassmann conditions only three can be satisfied for some configurationsof the MSSM namely :

11

MSSMC

y

xA1

A2

A4

B1

B3

B5

A1

A2

A4

mobile

base

Figure 8: The MSSM parallel manipulator (perspective and top view).

• 3d case : 4 lines are coplanar

• 5a case : all the 6 lines belong to a general complex

• 5b case : all the 6 lines intersect a line of space (special complex)

In this analysis we assume that it is not possible that a link lie in the baseplane.

6.1 Case 3d

We consider a set of 4 lines S and investigate for which configurations of themobile plate they can belong to a plane P. First we notice that the set of linesmust be such that the number of their distinct articulation points Ai is less thanthree. Indeed in the opposite case the plane P will be defined by the points(A1, A2, A3) i.e. the plane P will be the base plane and therefore the links willlie on the base plane. Thus the only possible sets for S are (1,2,3,6), (2,3,4,5),(1,4,5,6). Let us notice that in each case the points B1, B3, B5 belong to P i.e.this plane is the mobile plane. By changing the base and relative frame we canconsider that all these sets can be reduced to the set (1,2,3,6) (Figure 9). Thecoordinates of the articulation points on the base in the reference frame are :

A6 = A1 =

xa0

ya0

0

A3 = A2 =

xa1

ya0

0

A5 = A4 =

0ya3

0

The coordinates of the articulation points on the mobile in the relative frameare :

B2 = B1 =

0y0

0

B3 = B4 =

x2

y2

0

B5 = B6 =

x3

y2

0

12

1 2

3

45

6

1

23

4

56

1

2

3

4

5

6

A4, A5

B3, B4

B5, B6B1, B2

A1, A6

A2, A3

A4, A5

B3, B4

B5, B6

A2, A3

B1, B2

A1, A6

A1, A6 A2, A3

A4, A5

B5, B6

B1, B2

B3, B4

A1, A6

y x

y1

x1

y1x1

y

x

Figure 9: Various frames which can be used to study the case 3d.

13

The coplanarity of lines (1,2,3,6) can be expressed by two equations:

(A1B1 ∧ A2B2).A3B3 = 0 (A1B1 ∧A2B2).A6B6 = 0 (9)

which constitute a system of two linear equations in yc, zc in which xc does notappear. The determinant ∆ of this system is:

∆ = − sin(θ) sin(ψ)(xa0 − xa1)2(−x3 + x2)(−y2 + y0) (10)

Therefore in the case where ∆ is not equal to zero we solve this linear systemand we get a singularity condition described by :

yc = H3d1(Ωc) zc = H3d2

(Ωc) ∀xc (11)

But in that case it is possible to show that lines (1,2) are colinear and lie inthe base plane (in that case we have A1B1 ∧A2B2 = 0).

The determinant ∆ may vanish for sin θ = 0 i.e. for θ = 0 or θ = π but inboth cases the only possible solution for the equations (9) is zc = 0 : thereforethe links lie in the base plane. The determinant ∆ may also vanish for sinψ = 0i.e. for ψ = 0 or ψ = π. If ψ = 0 both equations yield to:

zc = −(ya0 − yc) sin(θ)

cos(θ)(12)

If ψ = π both equations yield to:

zc =(ya0 − yc) sin(θ)

cos(θ)(13)

Therefore we get two others singularity conditions:

ψ = 0 zc = H3d3(yc, θ) ∀xc (14)

ψ = π zc = −H3d4(yc, θ) ∀xc (15)

An example of this kind of singular configurations is presented in Figure 10.We may notice that we get the singular configuration described by Hunt.

6.2 Case 5a

In this case all the 6 lines belong to a general complex. This means also thatevery lines of the complex which are coplanar must intersect the same point.Let us consider the lines of the pencils spanned by (1,6), (2,3),(4,5). These linesbelong to the complex. Among these lines consider the three lines D1, D2, D3

which lie on the base plane. If the 6 lines belong to a linear general complexthe lines D1, D2, D3 must intersect the same point M , whose coordinates are

14

1 2

3

45

6

1

23

4

56

1

2

3

4

5

6

A4, A5

B3, B4

B5, B6B1, B2

A1, A6

A2, A3

A4, A5

B3, B4

B5, B6

A2, A3

B1, B2

A1, A6

A1, A6 A2, A3

A4, A5

B5, B6

B1, B2

B3, B4

A1, A6

y x

y1

x1

y1x1

y

x

Figure 10: An example of singular configuration of type 3d for the MSSM: 4 links arecoplanar.

15

(x, y, 0). If vij denotes the normal vector to the pencil of lines spanned by linei, j we must have:

A1M.v16 = 0 (16)

A3M.v23 = 0 (17)

A5M.v45 = 0 (18)

These three equations are linear in term of x, y. We use the first two tocalculate these unknowns and put their values in the last equation which yieldto a constraint equation. This equation is of order 3 in zc, 2 in xc, yc. Thereforewe get three possible singularity conditions :

a3(xc, yc,Ωc)z3c + a2(xc, yc,Ωc)z

2c + a1(xc, yc,Ωc)zc + a0(xc, yc,Ωc) = 0 (19)

b2(xc, zc,Ωc)y2c + b1(xc, zc,Ωc)yc + b0(xc, zc,Ωc) = 0 (20)

c2(yc, zc,Ωc)x2c + c1(yc, zc,Ωc)xc + c0(yc, zc,Ωc) = 0 (21)

Figure 11 shows three examples of singular configurations of type 5a obtainedfor fixed xc, yc,Ωc.

An interesting point about equation (19) is that for θ = 0 or θ = π thecoefficients ai are :

a0 = a1 = a2 = 0

a3 = 2x2y0 cos(ψ − φ)(xa0y0 + x2ya3) (22)

Thus in this case we get a singular configuration for ψ = ±π2

whatever arexc, yc, zc : we find the singular configuration described by Fichter.

A particular case has to be considered : let us assume that in the set ofequations (16,17,18) there are only two independent equations which can beused to determine x, y. As a consequence the last equation will not yield to asingularity condition. But if we consider the two dependent equations we willget such a condition by writing that their determinant is equal to zero if thetwo equations are coherent. The determinants of equations (16,17), (16,18),(18,17). are first order polynomials in xc. Thus we are able to get xc and putits value back in the equations. It can then be shown that the two equationsare not coherent.

7 Case 5b

In that case the 6 lines intersect one line of space. A previous analysis hasshown that five lines may intersect one line of space in two cases :

• the intersection line is an edge of the mobile and four lines of the manip-ulator are coplanar : this is Hunt’s singular configuration

16

1111

2

3

4

56

1

2

3

5

4

6

x0 = y0 = 0z0=9.412

ψ = θ = φ = 40

6

1

5

x0 = y0 = 0z0 = −21.2178

ψ = φ = θ = 40

1

2

3

2

3

4

6

1

x0 = y0 = 0

z0=1.2056

ψ = θ = φ = 40

6

5

Figure 11: Three examples of singular configuration of type 5a for the MSSM.

17

• three lines are coplanar and the intersection line (D) is defined by thetwo articulation points which are not common to two of the coplanarlines. For example in Figure 12 lines (2,3,4) are coplanar and the line (D)defined by A4, B1 intersects (1,2,3,4,5).

A1A2

A4

4

3

(D)

1

B3,4

B1,2

B5,6

2

M

6

52,3,4 sont

coplanaires

Figure 12: In this example the links (2,3,4) are coplanar and the line (D) going throughA4, B1 intersects the lines (1,2,3,4,5).

Let us consider this last case. By rotating the mobile plate around the edgeof the mobile defined by the articulation points of the three coplanar lines (inthe example edge B1, B3) we may find a configuration where the last line (6 inthe example) will intersect (D) (Figure 13).

Such a configuration is fully determined by its set of coplanar lines. These3 coplanar lines must share only two articulation points on the base (in theopposite case the plane on which they lie will be the base plane). In the samemanner they must also share only two articulation points on the mobile; indeedif they share three points the plane will be the mobile plane and any line whichis not in this set and which has a common articulation point with one line ofthe set will therefore be in the plane; thus we get 4 coplanar lines and this casehas been considered

Therefore the only possible set of three coplanar lines is (1,2,3), (1,5,6),(2,3,4), (3,4,5), (4,5,6). As in the previous section we can choose the referenceand relative frame so that we have to consider only the set (1,2,3).

These frames are defined in Figure 14. The coordinates in the referenceframe of the articulation points on the base are :

A6 = A1 =

xa0

00

A3 = A2 =

xa1

00

A5 = A4 =

0ya3

0

18

A1A2

A4

4

3

(D)

1

B3,4

B1,2

B5,6

2

M

6

5

Figure 13: In this example the links (2,3,4) are coplanar and the line (D) going throughA4, B1 intersects the lines (1,2,3,4,5). By rotating the mobile around its edge B1, B3 line 6intersects (D) at point M .

12

3

4

5

6

A1, A6 A2, A3

A4, A5

B5, B6

B1, B2

B3, B4

A1, A6

y

x

xr

yr

Figure 14: Notation and frame for singularity 5b.

19

The coordinates in the relative frame of the articulation points on the mobileare :

B2 = B1 =

0y0

0

B3 = B4 =

x2

00

B5 = B6 =

x3

00

The coplanarity of lines 1, 2, 3 is defined by the equation:

(A1B1 ∧A2B2).A3B3 = 0 (23)

Then we can express that the line going through A5, B5 intersects the linegoing through A1, B3 by the equation:

A1B3.(OA5 ∧ OB5) + A5B5.(OA1 ∧ OB3) = 0 (24)

Equations (23)(24) constitute a linear system in yc, zc. The determinant ∆ ofthis system is:

∆ = sin(θ)(xa0 − xa1)(x2 − x3)(x2 sin(φ)2ya3 sin(ψ) cos(θ) −

x2 sin(φ)ya3 cos(ψ) cos(φ) − cos(φ)y0ya3 sin(ψ) cos(θ) sin(φ) +

xa0y0 sin(ψ) + cos(φ)2y0ya3 cos(ψ))

If the determinant is not equal to zero we get then two singularity conditions:

yc = H5b1(xc,Ωc) zc = H5b2(xc,Ωc) (25)

An example of such singular configuration is given in Figure 15.

The determinant may be equal to zero if sin θ = 0. For θ = 0 equa-tions (23)(24) are reduced to:

−zc(−xa1 + xa0)(−sin(ψ)x2 + cos(ψ)y0) = 0 (26)

zc(ya3cos(ψ) + xa0sin(ψ))(−x3 + x2) = 0 (27)

The case where zc = 0 means that all the lines lie on the base plate. The othercase is obtained for :

tanψ =y0

x2

= −ya3

xa0

This case can hold only for a specific geometry of the robot.

For θ = π equations (23)(24) are reduced to:

zc(−xa1 + xa0)(−sin(ψ − φ)x2 + y0cos(ψ − φ)) = 0 (28)

−zc(sin(ψ − φ)xa0 − cos(ψ − φ)ya3)(−x3 + x2) = 0 (29)

20

1

2

3

45

6

1

5A1

A2

A5

B1

B5

B3

A1

A2

B3

A5

B5

B1

Figure 15: An example of singular configuration of type 5b for the MSSM: lines 1,2,3 arecoplanar and line A5B5 intersects line A1B3.

the case where zc = 0 means that all the lines lie on the base plate. The othercase is obtained for :

tan(ψ − φ) =y0

x2

=ya3

xa0

As before this case can hold only for a specific geometry of the robot.

The last case where the determinant is equal to zero is obtained for :

(x2 sin(φ)2ya3 sin(ψ) cos(θ) − x2 sin(φ)ya3 cos(ψ) cos(φ) + xa0y0 sin(ψ) +

cos(φ)2y0ya3 cos(ψ) − cos(φ)y0ya3 sin(ψ) cos(θ) sin(φ)) = 0 (30)

which can be solved in ψ :

ψ = − arctan(ya3 cos(φ)(cos(φ)y0 − x2 sin(φ))

x2ya3 cos(θ) sin(φ)2 + xa0y0 − cos(φ)y0ya3 cos(θ) sin(φ)) (31)

Then we can get zc from equation (23) and xc from equation (24). Thereforewe get three singularity conditions :

ψ = H5b3(θ, φ) zc = H5b4(yc, θ, φ) xc = H5b5(yc, θ, φ) (32)

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A1 A2

A3

B2B1

B3

Cn1

Figure 16: A 3 d.o.f parallel wrist. The mobile plate rotates around a fixed ball and socketjoint R whose center is C.

8 Summary of the singularity conditions for a MSSM

case singularity conditions3d ψ = 0 zc = H3d3(yc, θ) ∀xc

ψ = π zc = −H3d4(yc, θ) ∀xc

5a∑i=3

i=0 ai(xc, yc,Ωc)zic = 0

∑i=2i=0 bi(xc, zc,Ωc)y

ic = 0

∑i=2i=0 ci(yc, zc,Ωc)x

ic = 0

θ = φ = 0 ψ = ±π2 ∀(Xc)

5b yc = H5b1(xc,Ωc) zc = H5b2(xc,Ωc)ψ = H5b3(θ, φ) zc = H5b4(yc, θ, φ) xc = H5b5(yc, θ, φ)

9 Analysis of a 3 d.o.f. parallel wrist

The purpose of this section is to show that our geometric approach can beapplied to manipulator with less than 6 d.o.f. We consider the 3 d.o.f. parallelwrist presented in Figure 16. The mobile plate is articulated on a ball andsocket joint R which is fixed with respect to the base. Three variable lengthlinks articulated at point Ai, Bi enable to control the orientation of the mobileplate. If C is the center of the ball and socket joint and ni the unit vector oflink i the articular velocity ρ is related to the angular velocity of the mobileplate ω by :

ρ = J−1ω (33)

where row i of matrix J−1 is defined by:

J−1i = ((CBi ∧ ni)) (34)

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The singular configurations of this wrist are obtained when the matrix J−1 issingular. Let us denote by Fc the force vector applied on the ball and socketjoint, τ the articular force vector and T = (F,M) an external wrench appliedon the mobile plate. We have :

(

F

M

)

= Jf

(

Fc

τ

)

(35)

where Jf is a 6x6 matrix defined by:

... n1 n2 n3

I3...

0... CB1 ∧ n1 CB2 ∧ n2 CB3 ∧ n3

(36)

where I3 is the 3x3 identity matrix. It is easy to see that J−1 and Jf have thesame determinant. Then we notice that the three first columns of Jf are thePlucker vectors of the lines crossing C and parallel to the axis of the referenceframe. The last three column are simply the Plucker vectors of the link. Wemay thus apply our geometrical approach to this 6 lines to find the singularconfigurations of the wrist.

10 Conclusion

We have described a geometrical approach to determine the singular config-urations of closed-loop manipulator. This approach is in general much moresimpler than the classical approach which use the determinant of a jacobianmatrix. Another advantage is that we get also a geometrical description of thesingular configurations.

A complete analysis of a parallel manipulator has been presented. Thegeometrical approach enables to determine all the relations between the posi-tion and orientation parameters of the mobile plate which define the singularconfigurations.

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